Given two integers X and Y. The task is to find two vertices of an isosceles triangle ABC(right-angled at B) which has one vertex at a point B(0, 0). And there is a rectangle with opposite sides (0, 0) and (X, Y). All the points of this rectangle are located inside or on the border of the triangle. Print 4 integers x1, y1, x2, y2, where A(x1, y1) and B(x2, y2).
Examples:
Input : X = 3, Y = 3
Output : 6 0 0 6
Input : X = -3, y = -2
Output : -5 0 0 -5
Approach :
Let Val = |x| + |y|. Then the first point is (Val * sign(x), 0) and the second point is (0, Val * sign(y)).
Let’s see how it works for x > 0 and y > 0. Other cases can be proved in a similar way.
We need to show, that (x, y) belongs to our triangle(including its borders). In fact (x, y) belongs to segment, connecting (x + y, 0) with (0, x + y). The line through (x + y, 0) and (0, x + y) is Y = – X + x + y. Using coordinates (x, y) in this equation proves our answer.
Below is the implementation of the above approach:
C++
// C++ program to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)#include <bits/stdc++.h>using namespace std;// Function to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)int Vertices(int x, int y){ // Required value; int val = abs(x) + abs(y); // print x1 and y1 cout << val * (x < 0 ? -1 : 1) << " 0 "; // print x2 and y3 cout << "0 " << val * (y < 0 ? -1 : 1);}// Driver codeint main(){ int x = 3, y = 3; // Function call Vertices(x, y); return 0;} |
Java
// Java program to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)class GFG{// Function to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)static void Vertices(int x, int y){ // Required value; int val = Math.abs(x) + Math.abs(y); // print x1 and y1 System.out.print(val * (x < 0 ? -1 : 1) + " 0 "); // print x2 and y3 System.out.print("0 " + val * (y < 0 ? -1 : 1));}// Driver codepublic static void main(String[] args){ int x = 3, y = 3; // Function call Vertices(x, y);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find two vertices of an # isosceles triangle in which there is # rectangle with opposite side (0, 0) and (x, y) # Function to find two vertices of an # isosceles triangle in which there is # rectangle with opposite side (0, 0) and (x, y) def Vertices(x, y) : # Required value; val = abs(x) + abs(y); # print x1 and y1 if x < 0 : x = -1 else : x = 1 print(val * x,"0",end = " "); # print x2 and y3 if y < 0 : y = -1 else : y = 1 print("0",val * y); # Driver code if __name__ == "__main__" : x = 3; y = 3; # Function call Vertices(x, y); # This code is contributed by AnkitRai01 |
C#
// C# program to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)using System; class GFG{// Function to find two vertices of an// isosceles triangle in which there is// rectangle with opposite side (0, 0) and (x, y)static void Vertices(int x, int y){ // Required value; int val = Math.Abs(x) + Math.Abs(y); // print x1 and y1 Console.Write(val * (x < 0 ? -1 : 1) + " 0 "); // print x2 and y3 Console.Write("0 " + val * (y < 0 ? -1 : 1));}// Driver codepublic static void Main(String[] args){ int x = 3, y = 3; // Function call Vertices(x, y);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program to find two vertices of an // isosceles triangle in which there is // rectangle with opposite side (0, 0) and (x, y) // Function to find two vertices of an // isosceles triangle in which there is // rectangle with opposite side (0, 0) and (x, y) function Vertices(x, y) { // Required value; let val = Math.abs(x) + Math.abs(y); // print x1 and y1 document.write(val * (x < 0 ? -1 : 1) + " 0 "); // print x2 and y3 document.write("0 " + val * (y < 0 ? -1 : 1)); } // Driver code let x = 3, y = 3; // Function call Vertices(x, y); // This code is contributed by Surbhi Tyagi.</script> |
6 0 0 6
Time Complexity : O(1)
Auxiliary Space: O(1)
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